Let X be the number of material anomalies occurring in a particular region of an
ID: 3124813 • Letter: L
Question
Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodology for Probabilistic Life Prediction of Multiple-Anomaly Materials"† proposes a Poisson distribution for X. Suppose that = 4. (Round your answers to three decimal places.)
(a) Compute both
P(X 4)
and
P(X < 4).
P(X 4) =
P(X < 4) =
(b) Compute
P(4 X 6).
(c) Compute
P(6 X).
(d) What is the probability that the number of anomalies does not exceed the mean value by more than one standard deviation?
P(X 4) =
P(X < 4) =
Explanation / Answer
A)
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 4
x = the maximum number of successes = 4
Then the cumulative probability is
P(X<=4 ) = 0.629 [ANSWER]
*************
Note that P(fewer than x) = P(at most x - 1).
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 4
x = our critical value of successes = 4
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 3 ) = 0.43347012
Which is also
P(fewer than 4 ) = P(x<4) = 0.433 [ANSWER]
**************
b)
Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)
Here,
x1 = 4
x2 = 6
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 4
Then
P(at most 3 ) = 0.43347012
P(at most 6 ) = 0.889326022
Thus,
P(between x1 and x2) = 0.456 [ANSWER]
******************
c)
Note that P(at least x) = 1 - P(at most x - 1).
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 4
x = our critical value of successes = 6
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 5 ) = 0.785130387
Thus, the probability of at least 6 successes is
P(at least 6 ) = P(6<=x) = 0.215 [ANSWER]
*******************
d)
As
standard deviation = sqrt(u) = sqrt(4)=2
Then we mean the range 4-2<=x<=4+2 or 2<=x<=6.
Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)
Here,
x1 = 2
x2 = 6
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 4
Then
P(at most 1 ) = 0.091578194
P(at most 6 ) = 0.889326022
Thus,
P(between x1 and x2) = 0.797747827 [ANSWER]
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